Single Variable Calculus, Chapter 4, 4.5, Section 4.5, Problem 26

Use the guidelines of curve sketching to sketch the curve. $\displaystyle y = \frac{x}{\sqrt{x^2 -1} }$

The guidelines of Curve Sketching
A. Domain.
We know that $f(x)$ is a rational function that contains square root that is defined for positive values only. Hence,
$x^2 - 1 > 0 \\
x^2 > 1$
We have $x > 1$ and $x < -1$
Therefore, the domain is $(-\infty,-1)\bigcup(1,\infty)$

B. Intercepts.
Solving for $y$-intercept, when $x= 0$
$\displaystyle y = \frac{0}{\sqrt{0^2 - 1}} = 0$
Solving for $x$-intercept, when $y = 0$
$\displaystyle 0 = \frac{x}{\sqrt{x^2 -1 }}$
We have, $x = 0$ as $x$-intercept. However, 0 is not included in the domain therefore, we don't have intercepts

C. Symmetry.
Since $f(-x) = -f(x)$, the function is symmetric to the angle.

D. Asymptotes.
For vertical asymptotes, we set the denominator equal to 0.

$
\begin{equation}
\begin{aligned}
\sqrt{x^2 - 1 } &= 0\\
\\
x^2 & = 1\\
\\
x &= \pm 1
\end{aligned}
\end{equation}
$

For horizontal asymptotes,

$
\begin{equation}
\begin{aligned}
\lim_{x \to \pm\infty} \frac{x}{\sqrt{x^2 -1 }} &= \lim_{x \to \pm\infty} \frac{x}{\sqrt{x^2-1}} \frac{\left( \frac{1}{x}\right)}{\left( \frac{1}{x} \right)}\\
\\
&= \lim_{x \to \pm\infty} \frac{1}{\sqrt{1 - \frac{1}{x}}} = \pm 1
\end{aligned}
\end{equation}
$

Therefore, we have $y = \pm 1 $ as horizontal asymptotes.

E. Intervals of Increase or Decrease.
If we take the derivative of $f(x)$, by using Quotient Rule and Chain Rule.

$
\begin{equation}
\begin{aligned}
f'(x) &= \frac{\sqrt{x^2 -1}(1) - (x) \left( \frac{1}{2} (x^2 -1)^{\frac{-1}{2}} (2x) \right)}{(\sqrt{x^2 -1} )^2}\\
\\
f'(x) &= \frac{(x^2 -1)^{\frac{1}{2}} - \frac{x^2}{(x^2 -1)^{\frac{1}{2}}}}{x^2 -1 }= \frac{x^2 -1 - x^2}{(x^2 -1)^{\frac{3}{2}}} = \frac{-1}{(x^2 -1)^{\frac{3}{2}}}
\end{aligned}
\end{equation}
$

when $f'(x) = 0$
$\displaystyle 0 = \frac{-1}{(x^2 -1)^{\frac{3}{2}}}$
$f'(x) = 0$ does not exist. Thus,we don't have critical numbers.
Hence, the intervals of increase or decrease are...

$
\begin{array}{|c|c|c|}
\hline\\
\text{Interval} & f'(x) & f\\
\hline\\
x < -1 & - & \text{decreasing on } (-\infty, -1)\\
\hline\\
x > 1 & + & \text{decreasing on } (1,\infty)\\
\hline
\end{array}
$


F. Local Maximum and Minimum Values.
since $f'(x)$ doesn't change sign, we can say that the function has no local maximum or minimum.

G. Concavity and Points of Inflection.

$
\begin{equation}
\begin{aligned}
\text{if } f'(x) &= \frac{-1}{(x^2-1)^{\frac{3}{2}}} \text{, then}\\
\\
f''(x) &= \frac{(x^2-1)^{\frac{3}{2}} (0) - (-1) \left(\frac{3}{2} (x^2-1)^{\frac{1}{2}} (2x) \right)}{\left[ (x^2-1)^{\frac{3}{2}} \right]^2}\\
\\
f''(x) &= \frac{3x(x^2-1)^{\frac{1}{2}}}{(x^2-1)^3} = \frac{3x}{(x^2-1)^{\frac{5}{2}}}
\end{aligned}
\end{equation}
$


when $f''(x) = 0$,

$
\begin{equation}
\begin{aligned}
0 &= \frac{3x}{(x^2-1)^{\frac{5}{2}}}\\
\\
0 & = 3x
\end{aligned}
\end{equation}
$

We have, $x=0$ as our inflection point. However, it is not included in the domain. Therefore, we don't have inflection point.

Thus, the concavity in the interval of the domain is...

$
\begin{array}{|c|c|c|}
\hline\\
\text{Interval} & f''(x) & \text{Concavity}\\
\hline\\
x < -1 & - & \text{Downward}\\
\hline\\
x > 1 & + & \text{Upward}\\
\hline
\end{array}
$


H. Sketch the Graph.